18)Angel A and angle b are vertical angles so they are congruent angles. If angle A is (4x+5)° and angle B (-2(x-17.5))° Find th
e measure of angle B
2 answers:
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Answer:
Step-by-step explanation:
In order to find the max height the ball reached, we have to complete the square on that quadratic. That will also, conveniently so, give us the number of seconds it will take the ball to reach that max height, that answer to part b. Let's begin to complete the square. Normally, you would move the constant over to the other side of the equals sign, but there is no constant here. The next step is to get the leading coefficient to be a 1, and ours right now is a -15. So we have to factor it out. Here's where we start the process of completing the square.
Next step is to take half the linear term, square it, and add it to both sides. Our linear term is 20/3. Half of 20/3 is 20/6, and 20/6 squared is 400/36.
Because this is an equation, what we add to the left side also has to be added to the right. BUT we didn't just add in 400/36, because we have that -15 out front as a multiplier that refuses to be ignored. What we actually added in was -15(400/36):
The reason we do this is to create a perfect square binomial on the left which will serve as the number of seconds, h, in the vertex (h, k), where h is the number of seconds it takes the ball to reach its max height, k. Simplifying both sides then gives us:
Finally, we will move the right side over by the left and set the quadratic back equal to h(t):
and from that you can determine that the vertex is
.
The answer to a. is vound in the second number of our vertex: k, the max height. The max of the golf ball was 500/3 feet or 166 2/3 feet.
Part b is found in the first number of the vertex: h, the number of seconds it took the golf ball to reach that max height. The time it took was 3 1/3 seconds.
Part c. is to state the domain (the time) and the range (the height) of the ball.
Domain is
D: {x | 0 ≤ x ≤ 3 1/3} and
Range is
R: {y | 0 ≤ y ≤ 166 2/3}
A function 
is periodic if there is some constant 
such that 
for all 
in the domain of . Then is the "period" of .
Example:
If
, then we have 
, and so 
is periodic with period 
.
It gets a bit more complicated for a function like yours. We're looking for such that
Expanding on the left, you have
and
It follows that the following must be satisfied:
The first two equations are satisfied whenever
, or more generally, when 
and 
(i.e. any multiple of 4).
The second two are satisfied whenever
, and more generally when 
with (any multiple of 10/7).
It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.
Let's verify:
More generally, it can be shown that
is periodic with period
.
Example:
If
It gets a bit more complicated for a function like yours. We're looking for
Expanding on the left, you have
and
It follows that the following must be satisfied:
The first two equations are satisfied whenever
The second two are satisfied whenever
It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when
Let's verify:
More generally, it can be shown that
is periodic with period
This is a problem where you should make an equation.
Let's say x is the first week. It would mean
First Week : x
Second Week : x + 5
Third Week : 2(x+5)
You need to add these up to make your equation. Your equation will look something like this :
Now you simplify.
That means they sold 104 Boxes in the First Week.
So they sold 109 Boxes the Second Week.
109 x 2 = 218
And they sold 218 Boxes the Third Week.
I hope I was useful!
Let's say x is the first week. It would mean
First Week : x
Second Week : x + 5
Third Week : 2(x+5)
You need to add these up to make your equation. Your equation will look something like this :
Now you simplify.
That means they sold 104 Boxes in the First Week.
So they sold 109 Boxes the Second Week.
109 x 2 = 218
And they sold 218 Boxes the Third Week.
I hope I was useful!